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Gibbs Free Energy Calculation

  1. Jul 24, 2015 #1
    ΔG=ΔH-TΔSsystem and ΔSsystem=qp/T and isn't qp=ΔH? Which means ΔG=0 always which is obviously wrong.

    But the derivation for Gibbs Free Energy is based on qsurrounding=-qsystem so shouldn't the term be equals to 0?
     
  2. jcsd
  3. Jul 24, 2015 #2

    Ygggdrasil

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    ΔSsystem=qp/T is true only for reversible processes. What do you know about the ΔG of a reversible process?
     
  4. Jul 25, 2015 #3
    It is 0. So for a reversible process qp=ΔH and so ΔG=0?

    For an irreversible process how would the ΔS look like?

    Thanks
     
  5. Aug 7, 2015 #4
    Are you aware that ΔG and ΔS are functions only of the initial and final thermodynamic equilibrium states of a system, and are independent of path between the two states, whether reversible or irreversible? From what you have learned, do you know how to determine the change in S between two thermodynamic equilibrium states of a system? Are you familiar with the equation dG = -SdT + VdP, which applies to a differential change in temperature and pressure between two closely neighboring thermodynamic equilibrium states of a system?

    I think your real question is "how do one determine ΔG and ΔS between two thermodynamic equilibrium states of a system."

    Chet
     
  6. Aug 14, 2015 #5
    It is possible for ΔSsystem ≠ 0, even when qP = 0. Therefore, qP = 0, does not necessarily lead to ΔG = 0.

    P. Radhakrishnamurty
     
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